Show that any $f \in \mathbb{Q}(\langle x \rangle)$ is associate to any $\overline{f} \in \mathbb{Q}[x]$

48 Views Asked by At

Let $G = \langle x \rangle$ be the infinite cyclic group generated by an element $x$. The group ring $R = \mathbb{Q}(G)$ consists of finite sums of the form $$ r_{-m}x^{-m}+r_{-m+1}x^{-m+1}+\cdots + r_{-1}x^{-1} + r_{0} + r_{1}x + \cdots + r_{n}x^{n} $$

where $m,n \geq 0$ and $r_{-m}$, $r_{-m+1}$, $\cdots$, $r_{n}$ are rational numbers. The ring of polynomials $\mathbb{Q}[x]$ is naturally embedded in $R$.

I need to prove that any element $f \in R$ is an associate to the polynomial $\overline{f}\in \mathbb{Q}[x]$ (i.e., $f|\overline{f}$ and $\overline{f}|f$), the only thing is, I am not entirely sure what $\overline{f}$ is supposed to be in this context.

Is it some kind of conjugate? Is it just any polynomial in $\mathbb{Q}[x]$? (And so I'm just showing that any of the finite sums in $R$ is associate to any polynomial in $\mathbb{Q}[x]$?)

If someone could explain to me what $\overline{f}$ is supposed to be here, that would help immensely.

Thank you!